Editing Limiting reagent (section)

===Example===

20.0 g of [[iron (III) oxide]] (Fe<sub>2</sub>O<sub>3</sub>) are reacted with 8.00 g [[aluminium]] (Al) in the following [[thermite reaction]]:

:<chem>Fe2O3(s) + 2 Al(s) -> 2 Fe(l) + Al2O3(s)</chem>

Since the reactant amounts are given in grams, they must be first converted into moles for comparison with the chemical equation, in order to determine how many moles of Fe can be produced from either reactant.
* Moles of Fe which can be produced from reactant Fe<sub>2</sub>O<sub>3</sub>
*: <math chem>\begin{align}
 \ce{mol~Fe2O3} &= \frac{\ce{grams~Fe2O3}}{\ce{g/mol~Fe2O3}}\\
 &= \frac{20.0~\ce g}{159.7~\ce{g/mol}} = 0.125~\ce{mol}
\end{align}</math>
*: <math chem> \ce{mol~Fe} = 0.125 \ \ce{mol~Fe2O3} \times \frac{\ce{2~mol~Fe}}{\ce{1~mol~Fe2O3}} = 0.250~\ce{mol~Fe}</math>
* Moles of Fe which can be produced from reactant Al
*: <math chem>\begin{align}
 \ce{mol~Al} &= \frac\ce{grams~Al}\ce{g/mol~Al}\\
 & = \frac{8.00~\ce g}{26.98~\ce{g/mol}} = 0.297~\ce{mol}
\end{align}</math>
*: <math chem> \ce{mol~Fe} = 0.297~\ce{mol~Al} \times \frac\ce{2~mol~Fe}\ce{2~mol~Al} = 0.297~\ce{mol~Fe}</math>
There is enough Al to produce 0.297&nbsp;mol Fe, but only enough Fe<sub>2</sub>O<sub>3</sub> to produce 0.250&nbsp;mol Fe. This means that the amount of Fe actually produced is limited by the Fe<sub>2</sub>O<sub>3</sub> present, which is therefore the limiting reagent.